【HDU 4940】Destroy Transportation system（无源无汇带上下界可行流）

Description

Tom is a commander, his task is destroying his enemy’s transportation system.

Let’s represent his enemy’s transportation system as a simple directed graph G with n nodes and m edges. Each node is a city and each directed edge is a directed road. Each edge from node u to node v is associated with two values D and B, D is the cost to destroy/remove such edge, B is the cost to build an undirected edge between u and v.

His enemy can deliver supplies from city u to city v if and only if there is a directed path from u to v. At first they can deliver supplies from any city to any other cities. So the graph is a strongly-connected graph.

He will choose a non-empty proper subset of cities, let’s denote this set as S. Let’s denote the complement set of S as T. He will command his soldiers to destroy all the edges (u, v) that u belongs to set S and v belongs to set T.

To destroy an edge, he must pay the related cost D. The total cost he will pay is X. You can use this formula to calculate X:

After that, all the edges from S to T are destroyed. In order to deliver huge number of supplies from S to T, his enemy will change all the remained directed edges (u, v) that u belongs to set T and v belongs to set S into undirected edges. (Surely, those edges exist because the original graph is strongly-connected)

To change an edge, they must remove the original directed edge at first, whose cost is D, then they have to build a new undirected edge, whose cost is B. The total cost they will pay is Y. You can use this formula to calculate Y:

At last, if Y>=X, Tom will achieve his goal. But Tom is so lazy that he is unwilling to take a cup of time to choose a set S to make Y>=X, he hope to choose set S randomly! So he asks you if there is a set S, such that Y<X. If such set exists, he will feel unhappy, because he must choose set S carefully, otherwise he will become very happy.

Input

There are multiply test cases.

The first line contains an integer T(T<=200), indicates the number of cases.

For each test case, the first line has two numbers n and m.

Next m lines describe each edge. Each line has four numbers u, v, D, B.
(2=<n<=200, 2=<m<=5000, 1=<u, v<=n, 0=<D, B<=100000)

The meaning of all characters are described above. It is guaranteed that the input graph is strongly-connected.

Output

For each case, output "Case #X: " first, X is the case number starting from 1.If such set doesn’t exist, print “happy”, else print “unhappy”.

Sample Input

Sample Output

Case #1: happy

Case #2: unhappy

Hint

In first sample, for any set S, X=2, Y=4. In second sample. S= {1}, T= {2, 3}, X=10, Y=4.

Source

2014 Multi-University Training Contest 7

先来一个错误但是AC了(数据水)的算法：

每个点单独作为S集合时，如果存在满足Y<X，就输出unhappy。否则输出happy。

为什么错呢？输出happy时，即两个点单独作为S时，都有Y1>=X1,Y2>=X2，如果两个点之间没有边，它们一起作为S时，X=X1+X2,Y=Y1+Y2，则Y>=X；但是如果两个点有边相连，X=X1+X2-(S1到S2的D)-(S2到S1的D),Y=Y1+Y2-(S1到S2的D+B)-(S2到S1的D+B)，那么Y+B12+B21>=X,就有可能是Y<X了。当输入

1
4 5
1 4 10 2
4 1 6 2
2 1 1 1
4 3 6 2
3 2 4 2

时，输出的是happy。可是实际上，选择1、4节点，Y=2，X=8，显然是unhappy。

附上非正解代码：（AC了只能说明题目数据太水了）

#include <cstdio>

#include <cstring>

#define N 205

#define sf(x) scanf("%d",&x)

int x[N],y[N];

int main(){

    int t;

    sf(t);

    for(int cas=;cas<=t;cas++){

        memset(x,,sizeof x);

        memset(y,,sizeof y);

        printf("Case #%d: ",cas);

        int n,m;

        sf(n);

        sf(m);

        for(int i=;i<=m;i++){

            int u,v,d,b;

            sf(u);sf(v);sf(d);sf(b);

            x[u]+=d;

            y[v]+=d+b;

        }

        int ok=;

        for(int i=;i<=n;i++)

        if(x[i]>y[i])ok=;

        if(ok)puts("happy");

        else puts("unhappy");

    }

}

正解是无源无汇带上下界判断是否有可行流。

将问题转化为网络流问题：

每条边下界为D，上界为D+B，如果存在可行流，那么

$$\sum_{\substack{u\in S \\ v\in \overline {S}}} f_{uv} = \sum_{\substack{u\in S \\ v\in \overline {S}}}f_{uv}\\D_{uv} \leq f_{uv} \leq D_{uv}+B_{uv}$$

所以有

$$\sum_{\substack{u\in S \\ v\in \overline {S}}} D_{uv} \leq \sum_{\substack{u\in S \\ v\in \overline {S}}}D_{uv}+ B_{uv}$$

因此只要求无源无汇上下界网络流是否存在可行流，如果不存在就是unhappy。

而无源汇有上下界的网络流，是否有可行流可以这样求：

人为加上源点s，汇点t，

边权改为上界-下界（这样转化为下界为0），

流入i点的下界和为in，流出的下界和为out，

in>out则s 到 i 连边，流量为in-out；

in<out则 i 到 t 连边，流量为out-in。

求s到t的最大流，如果源点汇点连接的边全部满流则有可行解。

参考国家集训队论文《一种简易的方法求解流量有上下界的网络中网络流问题》

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstring>

#include <queue>

using namespace std;

#define N 500

#define M 200001

#define inf 0x3f3f3f3f

struct edge{

    int to,next,cap,flow;

}e[M];

int head[N],cnt;

int gap[N],dep[N],cur[N];

void init(){

    cnt=;

    memset(head, -, sizeof head);

}

void add(int u,int v,int w,int rw=){

    e[cnt]=(edge){v,head[u],w,};

    head[u]=cnt++;

    e[cnt]=(edge){u,head[v],rw,};

    head[v]=cnt++;

}

int q[N];

void bfs(int st,int ed){

    memset(dep,-,sizeof dep);

    memset(gap,,sizeof gap);

    gap[]=;

    int front=,rear=;

    dep[ed]=;

    q[rear++]=ed;

    while(front!=rear){

        int u=q[front++];

        for(int i=head[u];~i;i=e[i].next){

            int v=e[i].to;

            if(dep[v]!=-)continue;

            q[rear++]=v;

            dep[v]=dep[u]+;

            gap[dep[v]]++;

        }

    }

}

int s[N];

int sap(int st,int ed,int n){

    bfs(st,ed);

    memcpy(cur,head,sizeof head);

    int top=;

    int u=st;

    int ans=;

    while(dep[st]<n){

        if(u==ed){

            int Min=inf;

            int inser;

            for(int i=;i<top;i++)

                if(Min>e[s[i]].cap-e[s[i]].flow){

                    Min=e[s[i]].cap-e[s[i]].flow;

                    inser=i;

                }

            for(int i=;i<top;i++){

                e[s[i]].flow+=Min;

                e[s[i]^].flow-=Min;

            }

            ans+=Min;

            top=inser;

            u=e[s[top]^].to;

            continue;

        }

        bool flag=false;

        int v;

        for(int i=cur[u];~i;i=e[i].next){

            v=e[i].to;

            if(e[i].cap-e[i].flow&&dep[v]+==dep[u]){

                flag=true;

                cur[u]=i;

                break;

            }

        }

        if(flag){

            s[top++]=cur[u];

            u=v;

            continue;

        }

        int Min=n;

        for(int i=head[u];~i;i=e[i].next)

            if(e[i].cap-e[i].flow &&dep[e[i].to]<Min){

                Min=dep[e[i].to];

                cur[u]=i;

            }

        gap[dep[u]]--;

        if(!gap[dep[u]])return ans;

        gap[dep[u]=Min+]++;

        if(u!=st)u=e[s[--top]^].to;

    }

    return ans;

}

int main(){

    int t;

    scanf("%d",&t);

    for(int cas=;cas<=t;cas++){

        printf("Case #%d: ",cas);

        int n,m;

        scanf("%d%d",&n,&m);

        int in[N];

        int st=,ed=n+;

        memset(in,,sizeof in);

        init();

        for(int i=;i<=m;i++){

            int u,v,d,b;

            scanf("%d%d%d%d",&u,&v,&d,&b);

            add(u,v,b);

            in[v]+=d;

            in[u]-=d;

        }

        int need=;

        for(int i=;i<=n;i++){

            if(in[i]>){

                add(st,i,in[i]);

                need+=in[i];

            }

            else add(i,ed,-in[i]);

        }

        int ans=sap(st, ed, ed+);

        if(need==ans)puts("happy");

        else puts("unhappy");

    }

}